Which is the first non-zero digit from the right in the product of

can anybody explain it? thx!

We can convert \(4^3\) to \((2^2)^3 = 2^6\), similarly \(9^3 = 3^6\)

So now we have \(2^10\) and \(5^11\) we know 2*5=10, so \(2^10 * 5^10 = 10^10\)

Now only one 5 is left and whenever a 5 is multiplied with an odd number the last digit will always be a 5.

Hope this helps.

0s at the end are given by 10 as factors and combination of 2s and 5s (which again make 10s).

\(3^4 * 10^2\) will have 2 zeroes at the end.

\(3^4 * 2 * 5\) will have 1 zero at the end (one 2 and one 5 will multiply to give one 10)

In the given product, we have four 10s so four 0s at the end. We also have four 2s and eleven 5s which will make another four 10s. We do have more than four 5s but 5s alone cannot make a 10. You need 2s too and those are only four. So last eight digits of the number will be zeroes. Now, we have extra 5s leftover. Whenever we multiply a number by 5, it ends in 0 or 5 (depending on whether we have 2s or not). Since all 2s are already consumed, the rest of the number will end in 5. So the first non zero digit will be a 5.

for example:

\(3^2 * 2 * 5^2 * 10 = 9* 2*5 * 5 * 10 = 9* 10 * 5 * 10 = 45 * 100 = 4500\)


Answer (B)

Check this post for more on this concept : https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/0 ... actorials/

Are there not 2s in the 4s that will create more zeros when they multiply the extra 5s. In fact there seem to be six 2s available in the exponent of 4 that could multiply the extra 5s. i know the answer will still be B but i would love some clarification

5x2 will leave 0 therefore total power of 2 is 10 and total power of 5 is 11. one 5 is remaining and thereby last digit has to be 5.

Another analytical way to solve this : the product has 10 trailing zeros fomed by 10 5s and 10 2s, six of which come from 4^3. The product becomes 10^10 * something. We need to find the last digit of something. We then need to take the last digit of term in the product.The product is 5*3^7*7^2*9^3. Each term will have a last digit of 5,7,9 and 9 respectively.5*7*9*9 is 35 * 81. Once again we take the last digits of both terms which is 5 *1 . this gives us a last digit of 5.

Oh yes, most certainly! We have another 6 2s from 4^3 which I ignored. Thankfully, the 10 2s still don't add up to cancel out all 11 5s! So the last 5 will still hold down the fort and give us 5 as the first non zero digit.

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